Let us suppose we have a line $l_1 = ax + by + c = 0$ and we have a point P, $P(x_p, y_p)$ that is outside of this line. If we draw a perpendicular line from point $P$ to a point on the $l_1$. What would be the coordinates of this point ? I know there are many ways to do it. But I am looking for a single equation that can describe this point ($Q$) ?
Let $Q(x,y).$
Thus, since $PQ$ is perpendicular to our line $x$ and $y$ are solutions of the following system: $$ax+by+c=0$$ and $$bx-ay-bx_p+ay_p=0.$$ Can you end it now?
I got $$Q\left(\frac{b^2x_p-aby_p-ac}{a^2+b^2},\frac{a^2y_p-abx_p-bc}{a^2+b^2}\right).$$